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Transcript of Chapter Elecrical Systems
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1. Electrical Systems
1. ELECTRICAL SYSTEM
Syllabus
Electrical system: Electricity billing, Electrical load management and maximumdemand control, Power factor improvement and its benefit, Selection and location of
capacitors, Performance assessment of PF capacitors, Distribution and transformer losses.
1.1 Introduction to Electric Power Supply Systems
Electric power supply system in a country comprises of generating units that produce
electricity; high voltage transmission lines that transport electricity over long distances;
distribution lines that deliver the electricity to consumers; substations that connect the
pieces to each other; and energy control centers to coordinate the operation of the
components.
The Figure 1.1 shows a simple electric supply system with transmission anddistribution network and linkages from electricity sources to end-user.
Figure 1.1 Typical Electric Power Supply Systems
Power Generation Plant
The fossil fuels such as coal, oil and natural gas, nuclear energy, and falling water (hydel)
are commonly used energy sources in the power generating plant. A wide and growing
variety of unconventional generation technologies and fuels have also been developed,
including cogeneration, solar energy, wind generators, and waste materials.About 70 % of power generating capacity in India is from coal based thermal power
plants. The principle of coal-fired power generation plant is shown in Figure 1.2.
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Energy stored in the coal is converted in to electricity in thermal power plant. Coal is
pulverized to the consistency of talcum powder. Then powdered coal is blown into the
water wall boiler where it is burned at temperature higher than 1300oC. The heat in the
combustion gas is transferred into steam. This high-pressure steam is used to run the
steam turbine to spin. Finally turbine rotates the generator to produce electricity.
Figure 1.2 Principle of Thermal Power Generation
In India, for the coal based power plants, the overall efficiency ranges from 28% to 35%
depending upon the size, operational practices, fuel quality and capacity utilization.
Where fuels are the source of generation, a common term used is the HEAT RATE
which reflects the efficiency of generation. HEAT RATE is the heat input in kilo
Calories or kilo Joules, for generating one kilo Watt-hour of electrical output. One kilo
Watt hour of electrical energy being equivalent to 860 kilo Calories of thermal energy or3600 kilo Joules of thermal energy. The HEAT RATE expresses in inverse the
efficiency of power generation.
Transmission and Distribution Lines
The power plants typically produce 50 cycle/second
(Hertz), alternating-current (AC) electricity with
voltages between 11kV and 33kV. At the power plant
site, the 3-phase voltage is stepped up to a higher
voltage for transmission on cables strung on cross-
country towers.
High voltage (HV) and extra high voltage (EHV)
transmission is the next stage from power plant to
transport A.C. power over long distances at voltages
like; 220 kV & 400 kV. Where transmission is over
1000 kM, high voltage direct current transmission is
also favoured to minimize the losses.
Sub-transmission network at 132 kV, 110 kV, 66 kV or
33 kV constitutes the next link towards the end user.
Distribution at 11 kV / 6.6 kV / 3.3 kV constitutes thelast link to the consumer, who is connected directly or
through transformers depending upon the drawn level of
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service. The transmission and distribution network include sub-stations, lines and
distribution transformers. High voltage transmission is used so that smaller, more
economical wire sizes can be employed to carry the lower current and to reduce losses.
Sub-stations, containing step-down transformers, reduce the voltage for distribution to
industrial users. The voltage is further reduced for commercial facilities. Electricity must
be generated, as and when it is needed since electricity cannot be stored virtually in thesystem.
There is no difference between a transmission line and a distribution line except for the
voltage level and power handling capability. Transmission lines are usually capable of
transmitting large quantities of electric energy over great distances. They operate at high
voltages. Distribution lines carry limited quantities of power over shorter distances.
Voltage drops in line are in relation to the resistance and reactance of line, length and the
current drawn. For the same quantity of power handled, lower the voltage, higher the
current drawn and higher the voltage drop. The current drawn is inversely proportional to
the voltage level for the same quantity of power handled.
The power loss in line is proportional to resistance and square of current. (i.e. PLoss=I2R).
Higher voltage transmission and distribution thus would help to minimize line voltage
drop in the ratio of voltages, and the line power loss in the ratio of square of voltages.
For instance, if distribution of power is raised from 11 kV to 33 kV, the voltage drop
would be lower by a factor 1/3 and the line loss would be lower by a factor (1/3)2 i.e., 1/9.
Lower voltage transmission and distribution also calls for bigger size conductor on
account of current handling capacity needed.
Cascade Efficiency
The primary function of transmission and distribution equipment is to transfer power
economically and reliably from one location to another.
Conductors in the form of wires and cables strung on towers and poles carry the high-
voltage, AC electric current. A large number of copper or aluminum conductors are used
to form the transmission path. The resistance of the long-distance transmission conductors
is to be minimized. Energy loss in transmission lines is wasted in the form of I2R losses.
Capacitors are used to correct power factor by causing the current to lead the voltage.
When the AC currents are kept in phase with the voltage, operating efficiency of the
system is maintained at a high level.Circuit-interrupting devices are switches, relays, circuit breakers, and fuses. Each of
these devices is designed to carry and interrupt certain levels of current. Making and
breaking the current carrying conductors in the transmission path with a minimum of
arcing is one of the most important characteristics of this device. Relays sense abnormal
voltages, currents, and frequency and operate to protect the system.
Transformers are placed at strategic locations throughout the system to minimize
power losses in the T&D system. They are used to change the voltage level from low-to-
high in step-up transformers and from high-to-low in step-down units.
The power source to end user energy efficiency link is a key factor, which influences
the energy input at the source of supply. If we consider the electricity flow from
generation to the user in terms of cascade energy efficiency, typical cascade efficiencyprofile from generation to 11 33 kV user industry will be as below:
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Generation
Efficiency 1Efficiency ranges 28-35 % with respect to size of thermal
plant, age of plant and capacity utilization
Step-up Station2 Step-up to 400 / 800 kV to enable EHV transmission.Envisaged max. losses 0.5 % or efficiency of 99.5 %
EHV
Transmission &
Station 3
EHV transmission and substations at 400 kV / 800 kV.
Envisaged maximum losses 1.0 % or efficiency of 99 %
HV
Transmission &
Station 4
HV transmission & Substations for 220 / 400 kV.
Envisaged maximum losses 2.5 % or efficiency of 97.5 %
Sub-
transmission 5Sub-transmission at 66 / 132 kV
Envisaged maximum losses 4 % or efficiency of 96 %
Distribution
Station 6Step-down to a level of 11 / 33 kV.
Envisaged losses 0.5 % or efficiency of 99.5 %
Primary
Distribution 7Distribution is final link to end user at 11 / 33 kV.
Envisaged losses maximum 5 % of efficiency of 95 %
End user
Premises
Cascade efficiency from Generation to end user
= 1234567
The cascade efficiency in the T&D system from output of the power plant to the end use
is 87% (i.e. 0.995 x 0.99 x 0.975 x 0.96 x 0.995 x 0.95 = 87%)
Industrial End User
At the industrial end user premises, again the plant network elements like transformers at
receiving sub-station, switchgear, lines and cables, load-break switches, capacitors cause
losses, which affect the input-received energy. However the losses in such systems are
meager and unavoidable.
A typical plant single line diagram of electrical distribution system is shown in Figure 1.3
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MECHANICAL
ELECTRICAL
Figure 1.3 Electrical Distribution System Single Line Diagram
FANS & BLOWERS PUMPS R & AC COMPRESSORS LIGHTING, HEATING LOAD
M M M M C
M
M
DG SET
EB POWER SUPPLY
Metering
Transformer
.Effi.98-99.5%
Feeders,
Capacitors
Distribution
Panels
MotorsEffi.85-96%
TRIVECTOR METER
11 KV/440V
Transformer
C
C
Dist. loss
1 -6%Sub-station
ONE Unit saved = TWO Units Generated
After power generation at the plant it is transmitted and distributed over a wide network.The standard technical losses are around 17 % in India (Efficiency=83%). But the figures
for many of the states show T & D losses ranging from 17 50 %. All these may not
constitute technical losses, since un-metered and pilferage are also accounted in this loss.
When the power reaches the industry, it meets the transformer. The energy efficiency of
the transformer is generally very high. Next, it goes to the motor through internal plant
distribution network. A typical distribution network efficiency including transformer is
95% and motor efficiency is about 90%. Another 30 % (Efficiency=70%)is lost in the
mechanical system which includes coupling/ drive train, a driven equipment such as
pump and flow control valves/throttling etc. Thus the overall energy efficiency becomes
50%. (0.83 x 0.95x 0.9 x 0.70 = 0.50, i.e. 50% efficiency)
Hence one unit saved in the end user is equivalent to two units generated in the power
plant. (1Unit / 0.5Eff = 2 Units)
1.2 Electricity Billing
The electricity billing by utilities for medium & large enterprises, in High Tension (HT)
category, is often done on two-part tariff structure, i.e. one part for capacity (or demand)
drawn and the second part for actual energy drawn during the billing cycle. Capacity or
demand is in kVA (apparent power) or kW terms. The reactive energy (i.e.) kVArh drawn
by the service is also recorded and billed for in some utilities, because this would affect
the load on the utility. Accordingly, utility charges for maximum demand, active energy
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and reactive power drawn (as reflected by the power factor) in its billing structure. In addition,
other fixed and variable expenses are also levied.
The tariff structure generally includes the following components:
a) Maximum demand Charges
These charges relate to maximum demand registered during month/billing period and
corresponding rate of utility.
b) Energy Charges
These charges relate to energy (kilowatt hours) consumed during month / billing period
and corresponding rates, often levied in slabs of use rates. Some utilities now charge on
the basis of apparent energy (kVAh), which is a vector sum of kWh and kVArh.
c) Power factorpenalty or bonus rates, as levied by most utilities, are to contain reactive
power drawn from grid.
d) Fuel costadjustment charges as levied bysome utilities are to adjust the increasing fuel
expenses over a base reference value.
e) Electricity duty charges levied w.r.t units consumed.
f) Meter rentals
g) Lighting and fan power consumption is often at higher rates, levied sometimes on slab
basis or on actual metering basis.
h) Time Of Day (TOD) rates like peak and non-peak hours are also prevalent in tariff
structure provisions of some utilities.
i) Penalty for exceeding contract demand
j) Surcharge if metering is at LT side in some of the utilities
Analysis of utility bill data and monitoring its trends helps energy manager to identify ways for
electricity bill reduction through available provisions in tariff framework, apart from energy
budgeting.
The utility employs an electromagnetic or electronic trivector meter, for billing purposes. The
minimum outputs from the electromagnetic meters are
Maximum demand registered during the month, which is measured in preset timeintervals (say of 30 minute duration) and this is reset at the end of every billing cycle.
Active energy in kWh during billing cycle
Reactive energy in kVArh during billing cycle and
Apparent energy in kVAh during billing cycle
It is important to note that while maximum demand is recorded, it is not the instantaneous
demand drawn, as is often misunderstood, but the time integrated demand over the
predefined recording cycle.
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As example, in an industry, if the drawl over a recording cycle of 30 minutes is :
2500 kVA for 4 minutes
3600 kVA for 12 minutes
4100 kVA for 6 minutes
3800 kVA for 8 minutes
The MD recorder will be computing MD as:
(2500x4) +(3600 x 12) + (4100 x 6) + (3800 x 8) = 3606.7 kVA
30
The months maximum demand
will be the highest among such
demand values recorded over the
month. The meter registers only
if the value exceeds the previousmaximum demand value and
thus, even if, average maximum
demand is low, the industry /
facility has to pay for the
maximum demand charges for
the highest value registered
during the month, even if it
occurs for just one recording
cycle duration i.e., 30 minutes
during whole of the month. A typical
demand curve is shown in Figure1.4.
Figure 1.4 Demand Curve
As can be seen from the Figure 1.4 above the demand varies from time to time. The
demand is measured over predetermined time interval and averaged out for that interval
as shown by the horizontal dotted line.
Of late most electricity boards have changed over from conventional electromechanical
trivector meters to electronic meters, which have some excellent provisions that can help
the utility as well as the industry. These provisions include:
Substantial memory for logging and recording all relevant events High accuracy up to 0.2 class
Amenability to time of day tariffs
Tamper detection /recording
Measurement of harmonics and Total Harmonic Distortion (THD)
Long service life due to absence of moving parts
Amenability for remote data access/downloads
Trend analysis of purchased electricity and cost components can help the industry to
identify key result areas for bill reduction within the utility tariff available frameworkalong the following lines.
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Table 1.1 Purchased Electrical Energy Trend
Month&
Year
MDRecorded
kVA
BillingDemand*
kVA
Total EnergyConsumption
kWh
EnergyConsump-tion DuringPeak Hours
(kWh)
MDChargeRs./kVA
EnergyCharge
Rs./kWhPF
PFPenalty/Rebate Rs.
TotalBillsRs.
AverageCost
Rs./kWh
Jan.
Feb.
.
.
Dec.
*Some utilities charge Maximum Demand on the basis of minimum billing demand,
which may be between 75 to 100% of the contract demand or actual recorded demand
whichever is higher
1.3 Electrical Load Management and Maximum Demand Control
Need for Electrical Load Management
In a macro perspective, the growth in the electricity use and diversity of end use segments
in time of use has led to shortfalls in capacity to meet demand. As capacity addition is
costly and only a long time prospect, better load management at user end helps to
minimize peak demands on the utility infrastructure as well as better utilization of power
plant capacities.
The utilities (State Electricity Boards) use power tariff structure to influence end user in
better load management through measures like time of use tariffs, penalties on exceeding
allowed maximum demand, night tariff concessions etc. Load management is a powerful
means of efficiency improvement both for end user as well as utility.
As the demand charges constitute a considerable portion of the electricity bill, from user
angle too there is a need for integrated load management to effectively control the
maximum demand.
Step By Step Approach for Maximum Demand Control
1. Load Curve Generation
Presenting the load demand of a consumer
against time of the day is known as a load
curve. If it is plotted for the 24 hours of a
single day, it is known as a hourly load
curve and if daily demands plotted over a
month, it is called daily load curve. A
typical hourly load curve for an
engineering industry is shown in Figure1.5. These types of curves are useful in
Figure 1.5 Maximum Demand
(Daily Load Curve,Hourly kVA)
0
500
1000
1500
2000
2500
3000
KVA
7 8 9 10 11 12 13 14 15 16 17 18 1 9 20 21 22 23 24 1 2 3 4 5 6
Hours
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predicting patterns of drawl, peaks and valleys and energy use trend in a section or in an industry
or in a distribution network as the case may be.
2. Rescheduling of Loads
Rescheduling of large electric loads and equipment operations, in different shifts can be planned
and implemented to minimize the simultaneous maximum demand. For this purpose, it isadvisable to prepare an operation flow chart and a process chart. Analyzing these charts and with
an integrated approach, it would be possible to reschedule the operations and running equipment
in such a way as to improve the load factor which in turn reduces the maximum demand.
3.Storage of Products/in process material/ process utilities like refrigeration
It is possible to reduce the maximum demand by building up storage capacity of products/
materials, water, chilled water / hot water, using electricity during off peak periods. Off peak hour
operations also help to save energy due to favorable conditions such as lower ambient temperature
etc.
Example: Ice bank system is used in milk & dairy industry. Ice is made in lean period and used in
peak load period and thus maximum demand is reduced.
4. Shedding of Non-Essential Loads
When the maximum demand tends to reach preset limit, shedding some of non-essential loads
temporarily can help to reduce it. It is possible to install direct demand monitoring systems, which
will switch off non-essential loads when a preset demand is reached. Simple systems give an
alarm, and the loads are shed manually. Sophisticated microprocessor controlled systems are also
available, which provide a wide variety of control options like:
Accurate prediction of demand
Graphical display of present load, available load, demand limit
Visual and audible alarm
Automatic load shedding in a predetermined sequence
Automatic restoration of load
Recording and metering
5.Operation of Captive Generation and Diesel Generation Sets
When diesel generation sets are used to supplement the power supplied by the electric utilities, it
is advisable to connect the D.G. sets for durations when demand reaches the peak value. This
would reduce the load demand to a considerable extent and minimize the demand charges.
6. Reactive Power Compensation
The maximum demand can also be reduced at the plant level by using capacitor banks and
maintaining the optimum power factor. Capacitor banks are available with microprocessor based
control systems. These systems switch on and off the capacitor banks to maintain the desired
Power factor of system and optimize maximum demand thereby.
1.4 Power Factor Improvement and Benefits
Power factor Basics
In all industrial electrical distribution systems, the major loads are resistive and inductive.
Resistive loads are incandescent lighting and resistance heating. In case of pure resistive loads,the voltage(V),current (I), resistance (R)relations are linearly related,
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i.e.V = I R and Power (kW) = V I
Typical inductive loads are A.C. Motors, induction furnaces, transformers and ballast-
type lighting. Inductive loads require two kinds of power: a) active (or working) power to
perform the work and b) reactive power to create and maintain electro-magnetic fields.
Active power is measured in kW (Kilo Watts). Reactive power is measured in kVAr(Kilo Volt-Amperes Reactive).
The vector sum of the active power and reactive power make up the total (or apparent)
power used. This is the power generated by the SEBs for the user to perform a given
amount of work. Total Power is measured in kVA (Kilo Volts-Amperes) (See Figure 1.6).
Figure 1.7 Capacitors
The active power (shaft power required or true power required) in kW and the reactive
power required (kVAr) are 900 apart vectorically in a pure inductive circuit i.e., reactive
power kVAr lagging the active kW. The vector sum of the two is called the apparent
power or kVA, as illustrated above and the kVA reflects the actual electrical load on
distribution system.
Figure 1.6 kW, kVAr and kVA Vector
The ratio of kW to kVA is called the power factor, which is always less than or equal to
unity. Theoretically, when electric utilities supply power, if all loads have unity power
factor, maximum power can be transferred for the same distribution system capacity.
However, as the loads are inductive in nature, with the power factor ranging from 0.2 to
0.9, the electrical distribution network is stressed for capacity at low power factors.
Improving Power Factor
The solution to improve the power factor is to add power
factor correction capacitors (see Figure 1.7) to the plant
power distribution system. They act as reactive powergenerators, and provide the needed reactive power to
accomplish kW of work. This reduces the amount of
reactive power, and thus total power, generated by the
utilities.
Example:
A chemical industry had installed a 1500 kVA transformer. The initial demand of the
plant was 1160 kVA with power factor of 0.70. The % loading of transformer was about
78% (1160/1500 = 77.3%). To improve the power factor and to avoid the penalty, the unit
had added about 410 kVAr in motor load end. This improved the power factor to 0.89,and reduced the required kVA to 913, which is the vector sum of kW and kVAr (see
Figure 1.8).
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Figure 1.8 Power factor before and after ImprovementAfter improvement the plant had avoided penalty and the 1500 kVA transformer now
loaded only to 60% of capacity. This will allow the addition of more load in the future to
be supplied by the transformer.
The advantages of PF improvement by capacitor addition
a) Reactive component of the network is reduced and so also the total current in the
system from the source end.
b) I2R power losses are reduced in the system because of reduction in current.
c) Voltage level at the load end is increased.
d) kVA loading on the source generators as also on the transformers and lines upto
the capacitors reduces giving capacity relief. A high power factor can help in
utilising the full capacity of your electrical system.
Cost benefits of PF improvement
While costs of PF improvement are in terms of investment needs for capacitor addition
the benefits to be quantified for feasibility analysis are:
a) Reduced kVA (Maximum demand) charges in utility bill
b) Reduced distribution losses (KWH) within the plant network
c) Better voltage at motor terminals and improved performance of motors
d) A high power factor eliminates penalty charges imposed when operating with a low
power factor
e) Investment on system facilities such as transformers, cables, switchgears etc for
delivering load is reduced.
Selection and location of capacitors
Direct relation for capacitor sizing.
kVAr Rating = kW [tan 1 tan 2]
where kVAr rating is the size of the capacitor needed, kW is the average power drawn,
tan 1 is the trigonometric ratio for the present power factor, and tan 2 is the
trigonometric ratio for the desired PF.1 = Existing (Cos
-1 PF1) and 2 = Improved (Cos-1 PF2)
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Alternatively the Table 1.2 can be used for capacitor sizing.
The figures given in table are the multiplication factors which are to be multiplied with
the input power (kW) to give the kVAr of capacitance required to improve present power
factor to a new desired power factor.
Example: The utility bill shows an average power factor of 0.72 with an average KW of
627. How much kVAr is required to improve the power factor to .95 ?
Using formula
Cos 1 = 0.72 , tan 1 = 0.963Cos 2 = 0.95 , tan 2 = 0.329
kVAr required = P ( tan1 - tan2 ) = 627 (0.964 0.329)= 398 kVAr
Using table (see Table 1.2)
1) Locate 0.72 (original power factor) in column (1).
2) Read across desired power factor to 0.95 column. We find 0.635 multiplier
3) Multiply 627 (average kW) by 0.635 = 398 kVAr.
4) Install 400 kVAr to improve power factor to 95%.
Location of Capacitors
The primary purpose of capacitors is to
reduce the maximum demand. Additional
benefits are derived by capacitor location.
The Figure 1.6 indicates typical capacitor
locations. Maximum benefit of capacitors is
derived by locating them as close as
possible to the load. At this location, its
kilovars are confined to the smallest
possible segment, decreasing the load
current.
This, in turn, will reduce power losses of the
system substantially. Power losses are
proportional to the square of the current.When power losses are reduced, voltage at
the motor increases; thus, motor
performance also increases.
Locations C1A, C1B and C1C of Figure 1.9
indicate three different arrangements at the load. Note that in all three locations extra
switches are not required, since the capacitor is either switched with the motor starter or
the breaker before the starter. Case C1A is recommended for new installation, since the
maximum benefit is derived and the size of the motor thermal protector is reduced. In
Case C1B, as in Case C1A, the capacitor is energized only when the motor is in
operation. Case C1B is
Figure 1.9: Power Distribution Diagram Illustrating
Capacitor Locations
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Table 1.2 Multipliers To Determine Capacitor kVAr Requirements For Power Factor Correction
recommended in cases where the installation already exists and the thermal protector
does not need to be re-sized. In position C1C, the capacitor is permanently connected to
the circuit but does not require a separate switch, since capacitor can be disconnected by
the breaker before the starter.
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It should be noted that the rating of the capacitor should not be greater than the no-load
magnetizing kVAr of the motor. If this condition exists, damaging over voltage or
transient torques can occur. This is why most motor manufacturers specify maximum
capacitor ratings to be applied to specific motors.
The next preference for capacitor locations as illustrated by Figure 1.9 is at locations C2and C3. In these locations, a breaker or switch will be required. Location C4 requires a
high voltage breaker. The advantage of locating capacitors at power centres or feeders is
that they can be grouped together. When several motors are running intermittently, the
capacitors are permitted to be on line all the time, reducing the total power regardless of
load.
From energy efficiency point of view, capacitor location at receiving substation only
helps the utility in loss reduction. Locating capacitors at tail end will help to reduce loss
reduction within the plants distribution network as well and directly benefit the user by
reduced consumption. Reduction in the distribution loss% in kWh when tail end power
factor is raised from PF1 to a new power factor PF2, will be proportional to( )[ ]221 PF/PF-1 x 100
Capacitors for Other Loads
The other types of load requiring capacitor application include induction furnaces,
induction heaters and arc welding transformers etc. The capacitors are normally supplied
with control gear for the application of induction furnaces and induction heating furnaces.
The PF of arc furnaces experiences a wide variation over melting cycle as it changes from
0.7 at starting to 0.9 at the end of the cycle. Power factor for welding transformers is
corrected by connecting capacitors across the primary winding of the transformers, as the
normal PF would be in the range of 0.35.
Performance Assessment of Power Factor Capacitors
Voltage effects: Ideally capacitor voltage rating is to match the supply voltage. If the
supply voltage is lower, the reactive kVAr produced will be the ratio V12 /V2
2 where V1 is
the actual supply voltage, V2 is the rated voltage.
On the other hand, if the supply voltage exceeds rated voltage, the life of the capacitor is
adversely affected.
Material of capacitors: Power factor capacitors are available in various types bydielectric material used as; paper/ polypropylene etc. The watt loss per kVAr as well as
life vary with respect to the choice of the dielectric material and hence is a factor to be
considered while selection.
Connections: Shunt capacitor connections are adopted for almost all industry/ end user
applications, while series capacitors are adopted for voltage boosting in distribution
networks.
Operational performance of capacitors: This can be made by monitoring capacitor
charging current vis- a- vis the rated charging current. Capacity of fused elements can be
replenished as per requirements. Portable analyzers can be used for measuring kVArdelivered as well as charging current. Capacitors consume 0.2 to 6.0 Watt per kVAr,
which is negligible in comparison to benefits.
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Some checks that need to be adopted in use of capacitors are :
i) Nameplates can be misleading with respect to ratings. It is good to check by
charging currents.
ii) Capacitor boxes may contain only insulated compound and insulated terminalswith no capacitor elements inside.
iii) Capacitors for single phase motor starting and those used for lighting circuits for
voltage boost, are not power factor capacitor units and these cannot withstand
power system conditions.
1.5 Transformers
A transformer can accept energy at one voltage and deliver
it at another voltage. This permits electrical energy to be
generated at relatively low voltages and transmitted at highvoltages and low currents, thus reducing line losses and
voltage drop (see Figure 1.10).
Transformers consist of two or more coils that are
electrically insulated, but magnetically linked. The primary
coil is connected to the power source and the secondary coil
connects to the load. The turns ratio is the ratio between
the number of turns on the secondary to the turns on the
primary (See Figure 1.11).
The secondary voltage is equal to the primary voltage
times the turns ratio. Ampere-turns are calculated by
multiplying the current in the coil times the number of turns. Primary ampere-turns areequal to secondary ampere-turns. Voltage regulation of a transformer is the percent
increase in voltage from full load to no load.
Figure 1.10 View of a
Transformer
Types of Transformers
Transformers are classified as two categories: power
transformers and distribution transformers.
Power transformers are used in transmission network of
higher voltages, deployed for step-up and step down
transformer application (400 kV, 200 kV, 110 kV, 66 kV,
33kV)
Distribution transformers are used for lower voltage
distribution networks as a means to end user connectivity.
(11kV, 6.6 kV, 3.3 kV, 440V, 230V)Figure 1.11 Transformer
Rating of Transformer
Rating of the transformer is calculated based on the connected load and applying the
diversity factor on the connected load, applicable to the particular industry and arrive at
the kVA rating of the Transformer. Diversity factor is defined as the ratio of overall
maximum demand of the plant to the sum of individual maximum demand of various
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equipment. Diversity factor varies from industry to industry and depends on various
factors such as individual loads, load factor and future expansion needs of the plant.
Diversity factor will always be less than one.
Location of Transformer
Location of the transformer is very important as far as distribution loss is concerned.
Transformer receives HT voltage from the grid and steps it down to the required voltage.
Transformers should be placed close to the load centre, considering other features like
optimisation needs for centralised control, operational flexibility etc. This will bring
down the distribution loss in cables.
Transformer Losses and Efficiency
The efficiency varies anywhere between 96 to 99 percent. The efficiency of the
transformers not only depends on the design, but also, on the effective operating load.
Transformer losses consist of two parts: No-load loss and Load loss
1. No-load loss (also called core loss) is the power consumed to sustain the magnetic
field in the transformer's steel core. Core loss occurs whenever the transformer is
energized; core loss does not vary with load. Core losses are caused by two
factors: hysteresis and eddy current losses. Hysteresis loss is that energy lost by
reversing the magnetic field in the core as the magnetizing AC rises and falls and
reverses direction. Eddy current loss is a result of induced currents circulating in
the core.
2. Load loss (also called copper loss) is associated with full-load current flow in the
transformer windings. Copper loss is power lost in the primary and secondary
windings of a transformer due to the ohmic resistance of the windings. Copper
loss varies with the square of the load current. (P=I2R).
Transformer losses as a percentage of load is given in the Figure 1.12.
Figure 1.12 Transformer loss vs %Load
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For a given transformer, the manufacturer can supply values for no-load loss, PNO-LOAD,
and load loss, PLOAD. The total transformer loss, PTOTAL, at any load level can then be
calculated from:
PTOTAL = PNO-LOAD+ (% Load/100)2 x PLOAD
Where transformer loading is known, the actual transformers loss at given load can becomputed as:
)(
2
lossloadfullxkVARated
loadkVAlossloadNo
+=
Voltage Fluctuation Control
A control of voltage in a transformer is important due to frequent changes in supply
voltage level. Whenever the supply voltage is less than the optimal value, there is a
chance of nuisance tripping of voltage sensitive devices. The voltage regulation in
transformers is done by altering the voltage transformation ratio with the help of tapping.
There are two methods of tap changing facility available: Off-circuit tap changer and
On-load tap changer.
Off-circuit tap changerIt is a device fitted in the transformer, which is used to vary the voltage transformation
ratio. Here the voltage levels can be varied only after isolating the primary voltage of the
transformer.
On load tap changer (OLTC)
The voltage levels can be varied without isolating the connected load to the transformer.
To minimise the magnetisation losses and to reduce the nuisance tripping of the plant, the
main transformer (the transformer that receives supply from the grid) should be provided
with On Load Tap Changing facility at design stage. The down stream distribution
transformers can be provided with off-circuit tap changer.
The On-load gear can be put in auto mode or manually depending on the requirement.
OLTC can be arranged for transformers of size 250 kVA onwards. However, the
necessity of OLTC below 1000 kVA can be considered after calculating the cost
economics.
Parallel Operation of Transformers
The design of Power Control Centre (PCC) and Motor Control Centre (MCC) of any new
plant should have the provision of operating two or more transformers in parallel.
Additional switchgears and bus couplers should be provided at design stage.
Whenever two transformers are operating in parallel, both should be technically identical
in all aspects and more importantly should have the same impedance level. This will
minimise the circulating current between transformers.
Where the load is fluctuating in nature, it is preferable to have more than one transformerrunning in parallel, so that the load can be optimised by sharing the load between
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transformers. The transformers can be operated close to the maximum efficiency range by
this operation.
1.6 System Distribution Losses
In an electrical system often the constant no load losses and the variable load losses are to
be assessed alongside, over long reference duration, towards energy loss estimation.
Identifying and calculating the sum of the individual contributing loss components is a
challenging one, requiring extensive experience and knowledge of all the factors
impacting the operating efficiencies of each of these components.
For example the cable losses in any industrial plant will be up to 6 percent depending on
the size and complexity of the distribution system. Note that all of these are current
dependent, and can be readily mitigated by any technique that reduces facility current
load. Various losses in distribution equipment is given in the Table1.3.
In system distribution loss optimization, the various options available include:
Relocating transformers and sub-stations near to load centers
Re-routing and re-conductoring such feeders and lines where the losses /
voltage drops are higher.
Power factor improvement by incorporating capacitors at load end.
Optimum loading of transformers in the system.
Opting for lower resistance All Aluminum Alloy Conductors (AAAC) in place
of conventional Aluminum Cored Steel Reinforced (ACSR) lines
Minimizing losses due to weak links in distribution network such as jumpers,
loose contacts, old brittle conductors.
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1.7. Harmonics
Table 1.3 Losses in Electrical Distribution Equipment
% Energy Loss at Full
Load
Variations
S.No Equipment
Min Max
1. Outdoor circuit breaker (15 to 230 KV) 0.002 0.015
2. Generators 0.019 3.5
3. Medium voltage switchgears (5 to 15
KV)
0.005 0.02
4. Current limiting reactors 0.09 0.30
5. Transformers 0.40 1.90
6. Load break switches 0.003 0.025
7. Medium voltage starters 0.02 0.15
8. Bus ways less than 430 V 0.05 0.50
9. Low voltage switchgear 0.13 0.34
10. Motor control centers 0.01 0.4011. Cables 1.00 4.00
12. Large rectifiers 3.0 9.0
13. Static variable speed drives 6.0 15.0
14. Capacitors (Watts / kVAr) 0.50 6.0
In any alternating current network, flow of current depends upon the voltage applied and
the impedance (resistance to AC) provided by elements like resistances, reactances of
inductive and capacitive nature. As the value of impedance in above devices is constant,
they are called linear whereby the voltage and current relation is of linear nature.
However in real life situation, various devices like diodes, silicon controlled rectifiers,
PWM systems, thyristors, voltage & current chopping saturable core reactors, induction
& arc furnaces are also deployed for various requirements and due to their varying
impedance characteristic, these NON LINEAR devices cause distortion in voltage and
current waveforms which is of increasing concern in recent times. Harmonics occurs as
spikes at intervals which are multiples of the mains (supply) frequency and these distort
the pure sine wave form of the supply voltage & current.
Harmonics are multiples of the fundamental frequency of an electrical power system. If,
for example, the fundamental frequency is 50 Hz, then the 5th harmonic is five times that
frequency, or 250 Hz. Likewise, the 7th harmonic is seven times the fundamental or 350
Hz, and so on for higher order harmonics.
Harmonics can be discussed in terms of current or voltage. A 5th harmonic current is
simply a current flowing at 250 Hz on a 50 Hz system. The 5th harmonic current flowing
through the system impedance creates a 5th harmonic voltage. Total Harmonic Distortion
(THD) expresses the amount of harmonics. The following is the formula for calculating
the THD for current:
n
1
I
I
2100
2currentTHD
n n
n
=
=
=
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Then
%24100250
35
250
5022
=
+
= xITHD
When harmonic currents flow in a power system, they are known as poor power quality
or dirty power. Other causes of poor power quality include transients such as voltage
spikes, surges, sags, and ringing. Because they repeat every cycle, harmonics are regarded
as a steady-state cause of poor power quality.
When expressed as a percentage of fundamental voltage THD is given by,
THDvoltage = 1002
2
1
n
V
V
=
=
nn
n
where V1 is the fundamental frequency voltage and Vn is nth harmonic voltage
component.
Major Causes Of Harmonics
Devices that draw non-sinusoidal currents when a sinusoidal voltage is applied create
harmonics. Frequently these are devices that convert AC to DC. Some of these devices
are listed below:
Electronic Switching Power Converters
Computers, Uninterruptible power supplies (UPS), Solid-state rectifiers
Electronic process control equipment, PLCs, etc
Electronic lighting ballasts, including light dimmer
Reduced voltage motor controllers
Arcing Devices
Discharge lighting, e.g. Fluorescent, Sodium and Mercury vapor
Arc furnaces, Welding equipment, Electrical traction system
Ferromagnetic Devices
Transformers operating near saturation level
Magnetic ballasts (Saturated Iron core)
Induction heating equipment, Chokes, Motors
Appliances TV sets, air conditioners, washing machines, microwave ovens
Fax machines, photocopiers, printers
These devices use power electronics like SCRs, diodes, and thyristors, which are a
growing percentage of the load in industrial power systems. The majority use a 6-pulse
converter. Most loads which produce harmonics, do so as a steady-state phenomenon. A
snapshot reading of an operating load that is suspected to be non-linear can determine if it
is producing harmonics. Normally each load would manifest a specific harmonic
spectrum.
Many problems can arise from harmonic currents in a power system. Some problems are
easy to detect; others exist and persist because harmonics are not suspected. Higher RMS
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current and voltage in the system are caused by harmonic currents, which can result in
any of the problems listed below:
1. Blinking of Incandescent Lights - Transformer Saturation
2. Capacitor Failure - Harmonic Resonance
3. Circuit Breakers Tripping - Inductive Heating and Overload4. Conductor Failure - Inductive Heating
5. Electronic Equipment Shutting down - Voltage Distortion
6. Flickering of Fluorescent Lights - Transformer Saturation
7. Fuses Blowing for No Apparent Reason - Inductive Heating and Overload
8. Motor Failures (overheating) - Voltage Drop
9. Neutral Conductor and Terminal Failures - Additive Triplen Currents
10. Electromagnetic Load Failures - Inductive Heating
11. Overheating of Metal Enclosures - Inductive Heating
12. Power Interference on Voice Communication - Harmonic Noise
13. Transformer Failures - Inductive Heating
Overcoming Harmonics
Tuned Harmonic filters consisting of a capacitor bank and reactor in series are designed
and adopted for suppressing harmonics, by providing low impedance path for harmonic
component.
The Harmonic filters connected suitably near the equipment generating harmonics help to
reduce THD to acceptable limits. In present Indian context where no Electro Magnetic
Compatibility regulations exist as a application of Harmonic filters is very relevant for
industries having diesel power generation sets and co-generation units.
1.8Analysis of Electrical Power Systems
An analysis of an electrical power system may uncover energy waste, fire hazards, and
equipment failure. Facility /energy managers increasingly find that reliability-centered
maintenance can save money, energy, and downtime (see Table 1.4).
Table 1.4 Trouble shooting of Electrical Power SystemsSystem Problem Common Causes Possible Effects Solutions
Voltage imbalances
among the three
phases
Improper transformer tap
settings,
single-phase loads notbalanced among phases, poor
connections, bad conductors,
transformer grounds or faults.
Motor vibration,
premature motor failure
A 5% imbalance causes a
40% increase in motor
losses.
Balance loads amongphases.
Voltage deviations
from rated voltages
( too low or high)
Improper transformer settings, Over-voltages in motors
reduce efficiency, power
factor and equipment life
Incorrect selection of motors.
Increased temperature
Correct transformer
settings, motor ratings
and motor input
voltages
Poor connections in
distribution or atconnected loads.
Loose bus bar connections,
loose cable connections,
corroded connections, poorcrimps, loose or worn
contactors
Produces heat, causes
failure at connection site,
leads to voltage drops andvoltage imbalances
Use Infera Red
camera to locate hot-spots and correct.
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Undersized
conductors.
Facilities expanding beyond
original designs, poor power
factors
Voltage drop and energy
waste.
Reduce the load by
conservation load
scheduling.
Insulation leakage Degradation over time due toextreme temperatures,
abrasion, moisture, chemicals
May leak to ground or to
another phase. Variableenergy waste.
Replace conductors,insulators
Low Power Factor
Inductive loads such as
motors, transformers, and
lighting ballasts
Non-linear loads, such as
most electronic loads.
Reduces current-carrying
capacity of wiring,
voltage regulation
effectiveness, and
equipment life.
Add capacitors to
counteract reactive
loads.
Harmonics (non-
sinusoidal voltage
and/or current wave
forms)
Office-electronics, UPSs,
variable frequency drives,
high intensity discharge
lighting, and electronic andcore-coil ballasts.
Over-heating of neutral
conductors, motors,
transformers, switch gear.
Voltage drop, low powerfactors, reduced capacity.
Take care with
equipment selection
and isolate sensitive
electronics from noisycircuits.
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QUESTIONS
1. Name different types of power generation sources.
2. The temperatures encountered in power plant boilers is of the order ofa) 8500C b) 32000C c) 13000C d) 10000C
3. What do you understand by the term Heat Rate?
4. Explain why power is generated at lower voltage and transmitted at higher
voltages.
5. The efficiency of steam based power plant is of the order of
a) 28-35% b) 50-60% c) 70-75% d) 90-95%
6. The technical T & D loss in India is estimated to bea) 50% b) 25% c) 17% d) 10%
7. What are the typical billing components of the two-part tariff structure of
industrial utility?
8. Define contract demand and billing demand.
9. What are the areas to be looked into for maximum demand reduction in
industry?
10. A trivector-meter with half-hour cycle has the following inputs during themaximum demand period:
MD Drawn Duration
kVA in Minutes
100 10
200 5
50 10
150 5
What is the maximum demand during the half-hour interval?
11. Power factor is the ratio of
a) kW/kVA b) kVA/kW c) kVAr/kW d) kVAr/kVA
12. A 3-phase, 415 V, 100 kW induction motor is drawing 50 kW at a 0.75 PF
Calculate the capacitor rating requirements at motor terminals for
improving PF to 0.95. Also calculate the reduction in current drawn and
kVA reduction, from the point of installation back to the generated side due
to the improved PF.
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13. A process plant consumes of 12500 kWh per month at 0.9 Power Factor
(PF). What is the percentage reduction in distribution losses per month if
PF is improved up to 0.96 at load end?
14. What is the % loss reduction, if a 11 kV supply line is converted into 33 kV
supply system for the same length and electrical load application?
15. The efficiency at various stages from power plant to end-use is given
below. Efficiency of power generation in a power plant is 30 %. The T &
D losses are 23 %. The distribution loss of the plant is 6 %. Equipment
end use efficiency is 65 %. What is the overall system efficiency from
generation to end-use?
16. A unit has a 2 identical 500 kVA transformers each with a no load loss of
840 W and full load copper loss of 5700 watt. The plant load is 400 kVA.
Compare the transformer losses when single transformer is operation and
when both transformers are in parallel operation.
17. Explain how fluctuations in plant voltage can be overcome.
18. What are Total Harmonic Distortion and its effects on electrical system?
19. What are the equipments / devices contributing to the harmonics?
20. Select the location of installing capacitor bank, which will provide the
maximum energy efficiency.
a) Main sub-station b) Motor terminals c) Motor control centers d)
Distribution board
21. The designed power transformers efficiency is in the range of
a) 80 to 90.5 % b) 90 to 95.5 % c) 95 to 99.5 % d) 92.5 to 93.5 %
22. The power factor indicated in the electricity bill is
a) Peak day power factor b) Power factor during night c) Average power
factor d) Instantaneous power factor
REFERENCES
1. Technology Menu on Energy Efficiency NPC
2. NPC In-house Case Studies
3. Electrical energy conservation modules of AIP-NPC, Chennai